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Last of a series (correction of previous post)

by Loren Booda
Tags: correction, series
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Loren Booda
#1
Apr15-03, 12:51 AM
Loren Booda's Avatar
P: 3,408
Would you find the values of

[oo]
[sum] (-1)ncos(((2n)!)1/(2n))
n=1

and

[oo]
[sum] (-1)n+1sin(((2n+1)!)1/(2n+1))
n=0

for me? I am computer-challenged.
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#2
Apr15-03, 04:00 AM
P: 291
Are you sure that they converge?

From a first glance, I can see no reason why they should.
bogdan
#3
Apr15-03, 06:56 AM
P: 192
The 2 sums don't converge...

bogdan
#4
Apr15-03, 07:17 AM
P: 192
Last of a series (correction of previous post)

Picture of partial sums for the cos sum...


Picture of partial sums for the sin sum...
Loren Booda
#5
Apr15-03, 11:24 AM
Loren Booda's Avatar
P: 3,408
What sort of hardware/software do you guys use to work out these problems? Could you suggest some simple, inexpensive ware for math?

I'm sticking with the two constants, "L" and "B," that bogdan confirmed previously.
bogdan
#6
Apr16-03, 09:18 AM
P: 192
A 486 PC...or something like that...It costs somewhere between 50 and 100 dollars...
Loren Booda
#7
Apr16-03, 11:30 AM
Loren Booda's Avatar
P: 3,408
Is there any easily learned, inexpensive yet versatile math software out there that I could run on my PC?
imathgeek
#8
Apr16-03, 11:42 AM
P: 6
Inexpensive depends on what you're willing to pay for it. If you are affiliated with a college you may get an educational version of the software that would be much, much cheaper than the full blown versions, even get the full blown version at an academic price.

Otherwise, I think you are going to pay full price and that is expensive. The other option is to attend the computer/tech shows that seem to make the circuit of civic centers during the summer. I have seen some good prices on math software if you can find it.

Good luck.

Ken


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